Question

Show that n^2-n is divisible by 2 for every real number n

Answer 1

$\bold\red{Hello:-}$

Here is the answer to your question.

Any positive integer is of the form 2q or 2q + 1, where q is some integer

When n = 2q,

n²+n=(2q)²+2q

=4q²+2q

= 2q(2q+1)

which is divisible by 2

when n=2q+1

n²-n=(2q+1)²+(2q+1)

= 4q²+4q+1+2q+1

=4q²+6q +2

= 2(2q²+3q+1)

which is divisible by 2

hence n²+n is divisible by 2 for every positive integer n

Here is the answer to your question.

Any positive integer is of the form 2q or 2q + 1, where q is some integer

When n = 2q,

n²+n=(2q)²+2q

=4q²+2q

= 2q(2q+1)

which is divisible by 2

when n=2q+1

n²-n=(2q+1)²+(2q+1)

= 4q²+4q+1+2q+1

=4q²+6q +2

= 2(2q²+3q+1)

which is divisible by 2

hence n²+n is divisible by 2 for every positive integer n